Quantitative Aptitude
AVERAGES MCQs
Averages
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Question 71.
In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
Answer: Option B. -> 67 kgs
 - Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70.
According to Arun's mohter, X <= 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average = 66 + 67 + 68 = 201 = 67 kg. 3 3
 - Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70.
According to Arun's mohter, X <= 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average = 66 + 67 + 68 = 201 = 67 kg. 3 3
Answer: Option A. -> 21
 - Let the total number of workers be x. then,
8000x = (12000 x 7) + 6000 (x – 7)
2000x = 4200
x = 21
 - Let the total number of workers be x. then,
8000x = (12000 x 7) + 6000 (x – 7)
2000x = 4200
x = 21
Answer: Option B. -> 11
 - Let the number of papers be x.
Then, 63x + 20 + 2 = 65x or 2x = 22 or x = 11
 - Let the number of papers be x.
Then, 63x + 20 + 2 = 65x or 2x = 22 or x = 11
Answer: Option B. -> 77 2/3
 - Average = (76 x 16) - (75 x 10) 6 = 1216 - 750 6 = 466 6 = 233 3 = 77 2 3
 - Average = (76 x 16) - (75 x 10) 6 = 1216 - 750 6 = 466 6 = 233 3 = 77 2 3
Answer: Option B. -> 36.5
 - Correct sum = (36 x 50 + 48 – 23) = 1825
Correct mean =
1825 50
= 36.5
 - Correct sum = (36 x 50 + 48 – 23) = 1825
Correct mean =
1825 50
= 36.5
Answer: Option C. -> 85 kg
 - Total weight increased = (8 x 2.5) kg = 20 kg
Weight of new person = (65 + 20) kg = 85 kg
 - Total weight increased = (8 x 2.5) kg = 20 kg
Weight of new person = (65 + 20) kg = 85 kg
Answer: Option A. -> 51.4
 - Let the required mean score be a
Then, 20 x 80 + 25 x 31 + 55 x a = 52 x 100
1600 + 775 + 55a = 5200
55a = 2825
a = 51.4
 - Let the required mean score be a
Then, 20 x 80 + 25 x 31 + 55 x a = 52 x 100
1600 + 775 + 55a = 5200
55a = 2825
a = 51.4
Answer: Option A. -> 0 , 9
 - Let the number be x.
Then,
x + x2 2 = 5x x2 - 9x = 0 x (x - 9) = 0 x = 0 or x = 9.
 - Let the number be x.
Then,
x + x2 2 = 5x x2 - 9x = 0 x (x - 9) = 0 x = 0 or x = 9.
Answer: Option A. -> 3 M x M
 - We have : a + b + c 3 = M or (a + b + c) = 3M. Now, (a + b + c)M2 = (3M)M2 = 9M2 a2 + b2 + c2 + 2 (ab + bc + ca) = 9M2 a2 + b2 + c2 = 9M2 Required mean = a2 + b2 + c2 3 = 9M2 3 = 3M2
 - We have : a + b + c 3 = M or (a + b + c) = 3M. Now, (a + b + c)M2 = (3M)M2 = 9M2 a2 + b2 + c2 + 2 (ab + bc + ca) = 9M2 a2 + b2 + c2 = 9M2 Required mean = a2 + b2 + c2 3 = 9M2 3 = 3M2
Answer: Option B. -> 2 years
 - Total age of 5 members, 3 years ago = (17 x 5) years = 85 years
Total age of 5 members now = (85 + 3 x 5) years = 100 years
Total age of 6 members now = (17 x 6) years = 102 years
Age of the baby = (102 – 100) years = 2 years
 - Total age of 5 members, 3 years ago = (17 x 5) years = 85 years
Total age of 5 members now = (85 + 3 x 5) years = 100 years
Total age of 6 members now = (17 x 6) years = 102 years
Age of the baby = (102 – 100) years = 2 years